3
$\begingroup$

Find a basis for the plane $x-2y+3z=0$ in $R^3$. Then find a basis for the intersection of that plane with the $xy$ plane.

Is there a proper/algebraic way of finding the basis of a plane?
Just by looking at it a basis could be $(2, 1, 0)$ because any multiple of that will give you $0$ when you substitute, but how do I find this without guessing?

would I use the same process when finding the basis of a line?

Any hints on how to figure out the second part of the question?

$\endgroup$
  • 3
    $\begingroup$ A plane through the origin is a 2D subspace, so needs 2 basis vectors. The intersection of the plane with the $xy$ plane will be a 1D subspace, with 1 basis vector. $\endgroup$ – David Mar 18 '16 at 1:06
  • 2
    $\begingroup$ You can read off the normal vector of your plane. It is $(1,-2,3)$. Now, find the space of all vectors that are orthogonal to this vector (which then is the plane itself) and choose a basis from it. OR (easier): put in any 2 values for x and y and solve for z. Then $(x,y,z)$ is a point on the plane. Do that again with another random sample. Then the two choices are linearly independent with high probability. Check it and you're done. $\endgroup$ – Friedrich Philipp Mar 18 '16 at 1:12
  • 1
    $\begingroup$ You can use the idea in this question. In your case you have only one equation which means you have two free variables. $\endgroup$ – Mhenni Benghorbal Mar 18 '16 at 1:28
4
$\begingroup$

One thing you can identify is that $z = \frac{2y-x}{3}$, then the points that are going to satisfy the equality will be of the form $$ \left(x, y, \frac{2y-x}{3}\right). $$ For you to be able to cover all of such points, you would need to have two different vectors satisfying above such that they are not a multiple of each other.

You can see that indeed we can decompose the above vector as $$ \left(x, y, \frac{2y-x}{3}\right)=x\left(1,0,-\frac{1}{3}\right)+y\left(0,1,\frac{2}{3}\right). $$ which gives you an obvious basis $$ (v_1,v_2)=\left(1,0,-\frac{1}{3}\right)+\left(0,1,\frac{2}{3}\right). $$ There are many different ways of constructing a vector given in the first characterization, which would result in how the basis vector are aligned with respect to each other.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.